Simplify using following identity :

$(a \pm b)(a^2 \mp ab + b^2) = a^3 \pm b^3$

(i) (2x + 3y)(4x² - 6xy + 9y²)

(ii) $\Big(3x - \dfrac{5}{x}\Big)\Big(9x^2 + 15 + \dfrac{25}{x^2}\Big)$

(i) Given,

⇒ (2x + 3y)(4x² - 6xy + 9y²)

⇒ (2x + 3y)[(2x)² - 2x × 3y + (3y)²]

Comparing above equation with $(a \pm b)(a^2 \mp ab + b^2) = a^3 \pm b^3$, we get :

a = 2x and b = 3y

a³ + b³ = (2x)³ + (3y)³ = 8x³ + 27y³.

Hence, (2x + 3y)(4x² - 6xy + 9y²) = 8x³ + 27y³.

(ii) Given,

$\Rightarrow \Big(3x - \dfrac{5}{x}\Big)\Big(9x^2 + 15 + \dfrac{25}{x^2}\Big) \\[1em] \Rightarrow \Big(3x - \dfrac{5}{x}\Big)\Big[(3x)^2 + 3x \times \dfrac{5}{x} + \Big(\dfrac{5}{x}\Big)^2\Big]$

Comparing above equation with $(a \pm b)(a^2 \mp ab + b^2) = a^3 \pm b^3$, we get :

a = 3x and b = $\dfrac{5}{x}$

$\Rightarrow a^3 - b^3 = (3x)^3 - \Big(\dfrac{5}{x}\Big)^3 \\[1em] = 27x^3 - \dfrac{125}{x^3}.$

Hence, $\Big(3x - \dfrac{5}{x}\Big)\Big(9x^2 + 15 + \dfrac{25}{x^2}\Big) = 27x^3 - \dfrac{125}{x^3}$.